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4 years ago

5

A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t

he tractor takes 2.5 seconds to react and applies the breaks at a -1.65 m/s/s coming to a stop. How far did the hayride move before coming to a stop?

1 answer:

nexus9112 [7]4 years ago

6 0

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m$

So, the hayride will cover a distance of 68.18 m.

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A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50

Lelechka [254]

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

$s=u_1t+\frac{gt^2}{2}$

$1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}$

$u_1=6.11 m/s$

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

$v^2-u^2=2as$

where s= height of upper floor above window

here u=0

$6.11^2=2\times 9.81\times s$

$s=\frac{6.11^2}{2\times 9.81}$

s=1.902 m

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3 years ago

How much heat is absorbed by a 71g iron skillet when its temperature rises from 11oC to 29oC?

diamong [38]

The right formula to use for this calculation is the heat capacity formula,
Heat absorbed, Q = MCT, Where
M = Mass of the substance = 71
C = Specific heat capacity for iron = 0.450 J/gc
T = Change in temperature = 29 - 11 = 18
Q = 71 * 0.450 *18 = 575.10
The amount of heat absorbed by the iron skillet is 575 J.

5 0

3 years ago

What are three key aerodynamics principles?

mafiozo [28]

Answer:

A

Explanation:

There are three basic forces in aerodynamics: acceleration, which moves an airplane forward; drag, which holds it back; and height, which keeps it airborne. Lift is generally explained by three theories: Bernoulli's principle, the Coanda effect, and Newton's third law of motion.

8 0

3 years ago

1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15

kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity= $\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h$

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 $m/s^2$

Distance covered by the car,s = 75 m

Initial velocity of the car , $v_i$ = 6 m/s

Final velocity of thre car , $v_f$ =?

Using third equation of motion:

$v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2$

$v_{f}=14.69 m/s$

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0

3 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago